How to get the start of an email address - c

Question

I have two strings, one with an email address, and the other is empty.
If the email adress is e.g. "abc123#gmail.com", I need to pass the start of the email address, just before the # into the second string. For example:
first string: "abc123#gmail.com"
second string: "abc123"
I've written a loop, but it doesn't work:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
char email[256] = "abc123#gmail.com";
char temp[256];
int i = 0;
while (email[i] != '#')
{
temp = strcat(temp, email[i]);
i++;
}
printf ("%s\n", temp);
system ("PAUSE");
return 0;
}
Basically, I took every time one char from the email address, and added it into the new string. For example if the new string has a on it, now I'll put b with it too using strcat....

Answer 1

Pointers. Firstly, strcat() returns a char pointer, which C can't cast as a char array for some reason (which I hear all C programmers must know). Secondly, the second argument to strcat() is supposed to be a char pointer, not a char.
Replacing temp = strcat(temp, email[i]); with temp[i] = email[i]; should do the trick.
Also, after the loop ends, terminate the string with a null character.
temp[i] = '\0';
(After the loop ends, i is equal to the length of your extracted string, so temp[i] is where the terminal should go.)

Answer 2

There are better ways to solve this problem (e.g. by finding the index of the # (by strcspn or otherwise) and doing a memcpy), but your method is very close to working, so we can just make a few small adjustments.
As others have identified, the problem is with this line:
temp = strcat(temp, email[i]);
Presumably, you are attempting to copy the character at the ith position of email into the corresponding position of temp. However, strcat is not the correct way to do so: strcat copies data from one char* to another char*, that is, it copies strings. You just want to copy a single character, which is exactly what = does.
Looking at it from a higher level (so that I don't just tell you the answer), you want to set the appropriate character of temp to the appropriate character of email (you will need to use i to index both email and temp).
Also, remember that strings in C have to be terminated by '\0', so you have to set the next character of temp to '\0' after you have finished copying the string. (On this line of thought, you should consider what happens if your email string doesn't have an # in it, your while loop will keep going past the end of the string email: remember that you can tell if you are at the end of a string by character == '\0' or just using character as a condition.)

Answer 3

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
char email[256] = "abc123#gmail.com";
char temp[256];
size_t i = 0;
#if 0
for (i=0; email[i] && email[i] != '#'; i++) {;}
/* at the end of the loop email[i] is either the first '#',
** or that of the terminating '\0' (aka as strlen() )
*/
#else
i = strcspn(email, "#" );
/* the return value for strcspn() is either the index of the first '#'
* or of the terminating '\0'
*/
#endif
memcpy (temp, email, i);
temp[i] = 0;
printf ("%s\n", temp);
system ("PAUSE");
return 0;
}
UPDATE: a totally different approach would be to do the copying inside the loop (I guess this was the OP's intention):
for (i=0; temp[i] = (email[i] == '#' ? '\0' : email[i]) ; i++) {;}

Answer 4

You may want to try using strtok()